Multiply the following complex numbers: $({4}) \cdot ({-4-3i})$
Solution: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({4}) \cdot ({-4-3i}) = $ $ ({4} \cdot {-4}) + ({4} \cdot {-3}i) + ({0}i \cdot {-4}) + ({0}i \cdot {-3}i) $ Then simplify the terms: $ (-16) + (-12i) + (0i) + (0 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ -16 + (-12 + 0)i + 0i^2 $ After we plug in $i^2 = -1$ , the result becomes $ -16 + (-12 + 0)i - 0 $ The result is simplified: $ (-16 - 0) + (-12i) = -16-12i $